I would like to form common tangent equation for the functions f(x) and g(x) on the attachement. First of all I tried to form tangent equation for simple functions as y=x^2+2x+2 and y=-x^2-6x-14 but didn't manage. I am confused by now and any help would be appreciated.
Maybe it would be enough to show the way to get tangent equation for y=x^2+2x+2 and y=-x^2-6x-14 by explaining step by step (not only operations)

Thanks in advance.

Aug 17th 2005, 10:22 AM

rgep

The tangent to y = f(x) at the point (x0,y0) is the line through (x0,y0) with slope f'(x0): so y - f(x0) = f'(x0)(x-x0). So the tangent to y = log(x) is y-log(x0) = (1/x0)(x-x0). Similarly, the tangent to y = f(x) at the point (x1,y1) is y - f(x1) = f'(x1)(x-x1). So the tangent to y = x^2/2e is y - x1^2/2e = (x1/e)(x-x1).

You want these to be the same line. So you need the slopes and intercepts to be the same: that is, x1^2/2e = 1/x0 (slope) and log(x0)-1 = x1^2/2e - x1^2/e. You have to solve these to find x0 and x1: that gives you the equation for the common tangent.

Aug 17th 2005, 10:26 AM

Cold

I think this is right.
1, y=x^2 + 2x + 2
2, y=-x^2 - 6x - 14

OK suppose that the point of tangency of 1, is located at x=a. We can now calculate the y coordinate as a^2 + 2a + 2

I feel sure I've made an error but I haven't time to sort it.
Hope it helps.

Aug 17th 2005, 12:22 PM

rgep

There's a further technique that works when one of the equations is of the form y = quadratic in x (which it is here of course). Let y - f(x1) = f'(x1)(x-x1) be the equation for the tangent to the first curve: write this as y=mx+c. The line y = mx+c is tangent to the curve y = g(x) = quadratic in x if and only if the equation mx + c = g(x), which is a quadratic in x, has a repeated root: that is, iff its discriminant (b^2-4ac) is zero.

Cold, I analyzed your solution but it's someway strange.
I tried another way and according to the math program it is right.
So,
y=x^2+2x+2
y=-x^2-6x-14
The tangent we are going to find must be the same line which touch both parabola at different point.
The First touching point is (x1;y1)
The Second touching point is (x2;y2)
In order to not make equations hard to understand, I do substitution:
a=X1
b=x2
Because slope of the line is the same, I can equate the first derivatives of the functions, which means:
2a+2=-2b-6
2a+2b=-8
a+b=-4
b=-4-a
Now I can declare y1 and y2 by using a
y1=a^2+2a+2
y2=-b^2-6b-14=-(-a-4)^2-6(-a-4)-14)=-a^2-2a-6
Now I can form tangent of the first parabola, which means:
y-y1=k(x-x1) where k is slope
y-(a^2+2a+2)=(2a+2)(x-a)
y-a^2-2a-2=(2a+2)x-2a^2-2a
y=(2a+2)x-a^2+2As far as here, it was easy way. The following seems to me little bit wierd but nevertheless it brought me to the finish. Maybe some of you can explain it.
I considered the second touching point (x2;y2) and did the following substitution:
y2=(2a+2)x2-a^2+2
Now I substituted x2 and y1 by using a to get an equation to find the value of a:
-a^2-2a-6=(2a+2)(-a-4)-a^2+2
2a^2+8a=0
a=0
a=-4
It appears that there are two common tangent:
IF a=0
y1=0^2+2*0+2=2
b=0-4=-4
y2=-(-4^2)-6*(-4)-14=-6
According to the formula x-x1/x2-x1=y-y1/y2-y1 I can substitute the values and get the equation y=2x+2
IF a=-4
y1=(-4)^2+2*(-4)+2=10
b=-(-4)-4=0
y2=-0^2-6*0-14=-14
According to the formula x-x1/x2-x1=y-y1/y2-y1 I can substitute the values and get the equation y=-6x-14

And FINALLY I add an attachment to show that the equations of the line are true.

Aug 17th 2005, 01:06 PM

Dustin

What program did you use to do the graph?

Aug 17th 2005, 01:16 PM

totalnewbie

Quote:

Originally Posted by Dustin

What program did you use to do the graph?

I think that you have no use if I say to you the program's name because it's name is Function (application file is function.exe and I dont even recall where I donwloaded it).